The lifespans of seals in a particular zoo are normally distributed. The average seal lives $13.8$ years; the standard deviation is $3.2$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a seal living longer than $7.4$ years.
Solution: $13.8$ $10.6$ $17$ $7.4$ $20.2$ $4.2$ $23.4$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $13.8$ years. We know the standard deviation is $3.2$ years, so one standard deviation below the mean is $10.6$ years and one standard deviation above the mean is $17$ years. Two standard deviations below the mean is $7.4$ years and two standard deviations above the mean is $20.2$ years. Three standard deviations below the mean is $4.2$ years and three standard deviations above the mean is $23.4$ years. We are interested in the probability of a seal living longer than $7.4$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the seals will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the seals will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $7.4$ years and the other half $({2.5\%})$ will live longer than $20.2$ years. The probability of a particular seal living longer than $7.4$ years is ${95\%} + {2.5\%}$, or $97.5\%$.